You are given an integer array nums
with the following properties:
-
nums.length == 2 * n
. -
nums
containsn + 1
unique elements. - Exactly one element of
nums
is repeatedn
times.
Return the element that is repeated n
times.
Example 1:
Input: nums = [1,2,3,3]
Output: 3
Example 2:
Input: nums = [2,1,2,5,3,2]
Output: 2
Example 3:
Input: nums = [5,1,5,2,5,3,5,4]
Output: 5
Constraints:
-
2 <= n <= 5000
-
nums.length == 2 * n
-
0 <= nums[i] <= 104
-
nums
containsn + 1
unique elements and one of them is repeated exactlyn
times.
SOLUTION:
class Solution:
def repeatedNTimes(self, nums: List[int]) -> int:
n = len(nums)
nums.sort()
i = 0
while i < n:
ctr = 1
while i < n - 1 and nums[i] == nums[i + 1]:
ctr += 1
i += 1
i += 1
if 2 * ctr == n:
return nums[i - 1]
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